Balance the following chemical equation: $ $ $\text{Fe} +$ $\text{H}_2\text{O} \rightarrow$ $\text{Fe}_3\text{O}_4 +$ $\text{H}_2$
Solution: There are $3 \text{ Fe}$ on the right and only $1$ on the left, so multiply $\text{Fe}$ by ${3}$ $ {3}\text{Fe} + \text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2 $ There are $4 \text{ O}$ on the right and only $1$ on the left, so multiply $\text{H}_2\text{O}$ by ${4}$ $ 3\text{Fe} + {4}\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2 $ That gives us $8 \text{ H}$ on the left and only $2$ on the right, so multiply $\text{H}_2$ by ${4}$ . (Since hydrogen is by itself on the right, it should be done at the end because you can give it a coefficient without affecting another element.) $ 3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + {4}\text{H}_2 $ The balanced equation is: $ 3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2 $